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Question: A 0.2 molar solution of formic acid is 3.2% ionized, its ionization constant is: A. \(9.6\times {{...

A 0.2 molar solution of formic acid is 3.2% ionized, its ionization constant is:
A. 9.6×10−39.6\times {{10}^{-3}}
B. 2.1×10−42.1\times {{10}^{-4}}
C. 1.25×10−61.25\times {{10}^{-6}}
D. 2.1×10−82.1\times {{10}^{-8}}

Explanation

Solution

Dissociation in chemistry and biochemistry is a general mechanism in which, typically in a reversible way, molecules detach or break into smaller components such as electrons, ions, or radicals. The dissociation constant is termed the acid ionization constant (Ka{{K}_{a}}) for an aqueous solution of a mild acid. Similarly, baseline ionization constant (Kb{{K}_{b}}) is the equilibrium constant for the reaction of a weak base with water. For any acid – base pair of conjugates, KaKb=Kw{{K}_{a}}{{K}_{b}}={{K}_{w}}.

Complete answer:
The ionization for a general weak acid, HA, can be written as follows:
HAaq→Haq++Aaq−H{{A}_{aq}}\to H_{aq}^{+}+A_{aq}^{-}
Because the acid is weak, an equilibrium expression can be written. An acid ionization constant (Ka) is the equilibrium constant for the ionization of the acid.
Ka=[[H+][A−][HA]]{{K}_{a}}=\left[ \dfrac{[{{H}^{+}}][{{A}^{-}}]}{[HA]} \right]
The acid ionization represents the fraction of the original acid that has been ionized in solution. Therefore, the numerical value of Ka{{K}_{a}} is a reflection of the strength of the acid. Weak acids with relatively higher Ka{{K}_{a}} values are stronger acids that have relatively lower Ka{{K}_{a}} values. Because strong acids are essentially 100% ionized, the concentration of the acid in the denominator is nearly zero and the Ka{{K}_{a}} value approaches infinity. For this reason, Ka{{K}_{a}} values are generally reported for weak acids only.
HCOOH+H2O→H3O++COOH−HCOOH+{{H}_{2}}O\to {{H}_{3}}{{O}^{+}}+COO{{H}^{-}}
C 0 0
C – Cα Cα Cα
Ka{{K}_{a}} = dissociation or ionization constant.
Ka=C2α2C(1−α){{K}_{a}}=\dfrac{{{C}^{2}}{{\alpha }^{2}}}{C(1-\alpha )}
Ka=Cα21−α{{K}_{a}}=\dfrac{C{{\alpha }^{2}}}{1-\alpha }
C = initial concentration of undissociated acid.
α = extent up to which HX is ionized into ions.
According to given question,
C = 0.2M
α = 3.2% or 0.032
HCOOH→H++COOH−HCOOH\to {{H}^{+}}+COO{{H}^{-}}
C (1– α) Cα Cα
Ionization Constant, Ka=[COOH−][H+][HCOOH]{{K}_{a}}=\dfrac{[COO{{H}^{-}}][{{H}^{+}}]}{[HCOOH]}
Ka=cα×cαc(1−α){{K}_{a}}=\dfrac{c\alpha \times c\alpha }{c(1-\alpha )}
=(0.2×0.032)20.2×0.968=\dfrac{{{\left( 0.2\times 0.032 \right)}^{2}}}{0.2\times 0.968}
= 2.1×10−42.1\times {{10}^{-4}}

Hence, option B is the correct answer.

Note:
Always remember, Weak acids with relatively higher Ka{{K}_{a}} values are stronger acids that have relatively lower Ka{{K}_{a}} values.
Don’t get confused between molarity and molality. Molarity (M) is defined as number of moles of solute dissolved in one litre of solution whereas Molality (m) is defined as number of moles of solute per kilogram (kg) of the solvent